检查给定范围内所有数字的异或是偶数还是奇数

给定一个范围[L，R]，任务是查找范围L到R(包括两个端点)中所有自然数的XOR值是偶数还是奇数。如果范围内所有数字的异或为偶数，则打印“偶数”，否则为奇数。
例子：

Input: L = 1, R= 10 
Output: Odd

Input: L= 5, R=15
Output: Even

一个简单的解决方案是计算范围为[L，R]的所有数字的XOR，然后检查所得的XOR值是偶数还是奇数。
这种方法的时间复杂度将为O(n)。
一个有效的解决方案基于以下事实：

odd ^ odd = even
odd ^ even = odd
even ^ odd = odd
even ^ even = even

所有偶数的XOR将是偶数(与范围大小无关)，并且如果奇数的计数是奇数，则最终XOR将是奇数，如果偶数则最终XOR将是偶数。
现在可以得出结论，

• If the count of Odd Numbers is even, 
  XOR of all odd numbers = Even 
  XOR of all even numbers = Even 
  Final XOR = Even ^ Even = Even
• If the count of Odd Numbers is Odd, 
  XOR of all odd numbers = Odd 
  XOR of all even numbers = Even 
  Final XOR = Odd ^ Even = Odd

因此，我们要做的就是计算L到R范围内的奇数。
方法 ：

• 计算[L，R]范围内的奇数。
• 检查奇数计数是偶数还是奇数。
• 如果计数为偶数则打印“偶数”，否则为“奇数”。

下面是上述方法的实现：

C++

// C++ program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
 
#include 
using namespace std;
 
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
 
string isEvenOrOdd(int L, int R)
{
    // Count odd Numbers in range [L, R]
    int oddCount = (R - L) / 2;
 
    if (R % 2 == 1 || L % 2 == 1)
        oddCount++;
 
    // Check if count of odd Numbers
    // is even or odd
 
    if (oddCount % 2 == 0)
        return "Even";
    else
        return "Odd";
}
 
// Driver Code
int main()
{
 
    int L = 5, R = 15;
 
    cout << isEvenOrOdd(L, R);
 
    return 0;
}

Java

// Java program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
 
class GFG {
 
    // Function to check if XOR of all numbers
    // in range [L, R] is Even or Odd
 
    static String isEvenOrOdd(int L, int R)
    {
        // Count odd Numbers in range [L, R]
        int oddCount = (R - L) / 2;
 
        if (R % 2 == 1 || L % 2 == 1)
            oddCount++;
 
        // Check if count of odd Numbers
        // is even or odd
 
        if (oddCount % 2 == 0)
            return "Even";
        else
            return "Odd";
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int L = 5, R = 15;
 
        System.out.println(isEvenOrOdd(L, R));
    }
}

C#

// C# program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
 
using System;
class GFG {
 
    // Function to check if XOR of all numbers
    // in range [L, R] is Even or Odd
 
    static string isEvenOrOdd(int L, int R)
    {
        // Count odd Numbers in range [L, R]
        int oddCount = (R - L) / 2;
 
        if (R % 2 == 1 || L % 2 == 1)
            oddCount++;
 
        // Check if count of odd Numbers
        // is even or odd
 
        if (oddCount % 2 == 0)
            return "Even";
        else
            return "Odd";
    }
 
    // Driver Code
    public static void Main()
    {
 
        int L = 5, R = 15;
 
        Console.WriteLine(isEvenOrOdd(L, R));
    }
}

Python3

# Python3 program to check if XOR of
# all numbers in range [L, R]
# is Even or odd
 
 
# Function to check if XOR of all numbers
# in range [L, R] is Even or Odd
 
def isEvenOrOdd( L, R ):
 
    # Count odd Numbers in range [L, R]
    oddCount = (R - L )/2
     
    if( R % 2 == 1 or L % 2 == 1):
        oddCount = oddCount + 1
     
     
    # Check if count of odd Numbers
    # is even or odd
     
    if(oddCount % 2 == 0 ):
        return "Even"
    else :
        return "Odd"
         
 
     
# Driver Code
 
L = 5
R = 15
 
print(isEvenOrOdd(L, R));

Javascript

// JavaScript program to check if XOR of
// all numbers in range [L, R]
// is Even or odd
 
 
// Function to check if XOR of all numbers
// in range [L, R] is Even or Odd
 
function isEvenOrOdd(L, R)
{
    // Count odd Numbers in range [L, R]
    let oddCount = Math.floor((R - L) / 2);
 
    if (R % 2 == 1 || L % 2 == 1)
        oddCount++;
 
    // Check if count of odd Numbers
    // is even or odd
 
    if (oddCount % 2 == 0)
        return "Even";
    else
        return "Odd";
}
 
// Driver Code
 
    let L = 5, R = 15;
 
    document.write(isEvenOrOdd(L, R));
 
// This code is contributed by Surbhi Tyagi.
 

输出：

Even

时间复杂度：O(1)
辅助空间：O(1)